For this question, all operators and states are on a finite dimensional Hilbert space.

Suppose I have a collection of continuously parametrized Hamiltonians $H(t), 0\leq t\leq T$. Suppose furthermore that I have a time-independent Hermitian operator $O$ such that $[H(t), O] = 0$ for all $t$. Informally, the adiabatic theorem states that if $|\psi(0)\rangle$ is an eigenstate of $H(0)$, then, provided the evolution of $H(t)$ is sufficiently slow/long, $|\psi(t)\rangle$ will remain an eigenstate of $H(t)$ for all $t$. I am wondering what I can say about the relationship between $|\psi(t)\rangle$ and the symmetry operator $O$. Suppose for instance that $|\psi(0)\rangle$ is also an eigenstate of $O$ with eigenvalue $\lambda$. My questions are

  1. Is it true that $|\psi(t)\rangle$ is an eigenstate of $O$ for all $t$? If so, will it necessarily have the same eigenvalue $\lambda$ as $|\psi(0)\rangle$?
  2. Does the physics change if the eigenspace of $O$ associated to $\lambda$ is degenerate?

1 Answer 1

  1. If $| \psi(0) \rangle$ is initially an eigenvalue of $O$ with eigenvalue $\lambda$, and $O$ commutes with your Hamiltonian $H(t)$ for all times $t$, then indeed the state will always be an eigenvalue of $O$ with eigenvalue $\lambda$. This follows trivially by writing the time evolution operator in terms of $H$. The most general form possible for the time evolution operator is obtained by the time ordered exponential $$ U(t, t_0) = T \exp \Bigg\{ -i \int_{t_0}^t dt' H(t') \Bigg\} = \sum_{n = 0}^{\infty} \frac{(-i)^n}{n!} \int_{t_0}^t dt_1 \ldots \int_{t_0}^t dt_n \ T \Big\{ H(t_1) \ldots H(t_n) \Big\}, $$ where $T$ is the time-ordering symbol, which orders the $n$ instances of the Hamiltonian from latest to earliest in time. This form for the time evolution operator is necessary when (for instance) $H$ does not commute with itself at different times. In any case, you can immediately see that $[O,H(t)] = 0$ implies $[O,U(t)] = 0$ (I've set $t_0 = 0$), and therefore $$ O |\psi(t) \rangle = O U(t) |\psi(0) \rangle = U(t) O | \psi(0) \rangle = \lambda U(t) | \psi(0) \rangle = \lambda | \psi(t) \rangle $$

  2. From the above, if the eigenspace associated to $\lambda$ is nondegenerate then clearly the state $| \psi(t) \rangle$ is proportional to $| \psi(0) \rangle$ at all times. On the other hand, if the eigenspace is degenerate then there's no reason $| \psi(t) \rangle$ can't explore the full eigenspace. As a trivial example, take $O = 1$ to be the identity operator on a 2-dimensional Hilbert space, and let $H(t)$ be anything that evolves nontrivially.


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